There is also $1$ way the number $1$ appears three times face down, and the fact that $27+27+9+1=64$ should be encouraging.
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HenryJan 20 '12 at 21:55

3

That sounds right. As a check, you should argue that $1$ can appear on the bottom on no rolls, one roll, two rolls, or three rolls (there are no other possibilities), and you have worked out the number of ways that $1$ never appears, appears once, and appears twice for a total of $27 + 27 + 9 = 63$ of the $64$ occurrences. The remaining possibility is that of $1$ on all three rolls, right? Of course, all the numbers adding up doesn't guarantee that they are correct, but if they don't add up, that suggests that the work and the argument needs to be checked.
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Dilip SarwateJan 20 '12 at 21:57

@Dilip You should make that an answer.
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David MitraJan 20 '12 at 21:59

1 Answer
1

An alternate/complimentary way to answer question 4 is:
If 2 throws are '1' then we have only one throw which is not '1'. There are 3 ways for the not-'1' throw to be put down. And not-'1' has 3 cases as you said: {2,3,4}. So again $3 \cdot 3 = 9$

This might make you notice that in general: C(N,1) = C(N, N-1) = N . In other words I can choose 1 from N, or complimentary (and exactly equivalently) I can choose N-1 from N. So if we are throwing the tetrahedron N times you could either ask: 1) how many ways to roll '1' (N-1) times?, or 2) how many ways to roll not-'1' one time? It's the same question isn't it? And combinatorics confirms it.

And more general C(N, i) = C(N, N-i) You might already know that, but maybe now you have more insight into it.

Can you give the general solution to you problem? That is, let's say we have a set of N numbers (a dice of N sides) and we throw it M times. How many possible ways exist to get the same number K times?